Question

Resolva as equações de 2 grau.​

Answer 1

Resposta:

Explicação passo-a-passo:

a) x²-3x-4=0

a=1; b= -3 e c= -4

$\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(-4)=9-(-16)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{25}}{2(1)}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{25}}{2(1)}=\frac{3+5}{2}=\frac{8}{2}=4$

b) x²+6x+9=0

a=1; b=6 e c=9

$\Delta=(b)^{2}-4(a)(c)=(6)^{2}-4(1)(9)=36-(36)=0\\\\x^{'}=x^{''}= \frac{-(b)\pm\sqrt{\Delta}}{2(a)}=\frac{-(6)\pm\sqrt{0}}{2(1)}=\frac{-6\pm0}{2}=\frac{-6}{2}=-3$

c) x²-8x+15=0

a=1; b= -8 e c=15

$\Delta=(b)^{2}-4(a)(c)=(-8)^{2}-4(1)(15)=64-(60)=4\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-8)-\sqrt{4}}{2(1)}=\frac{8-2}{2}=\frac{6}{2}=3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-8)+\sqrt{4}}{2(1)}=\frac{8+2}{2}=\frac{10}{2}=5$

d) x²+x-12=0

a=1; b=1 e c= -12

$\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-12)=1-(-48)=49\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{49}}{2(1)}=\frac{-1-7}{2}=\frac{-8}{2}=-4\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{49}}{2(1)}=\frac{-1+7}{2}=\frac{6}{2}=3$